Section 5.4 More Trigonometric Graphs

Section 5.4 More Trigonometric Graphs
Graphs of the Tangent, Cotangent, Secant, and Cosecant Function
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REMARK: Many curves have a “U” shape near zero. For example, notice that the functions
sec x and x2 + 1 are very different functions. But when we casually sketch their graphs near
x = 0 the sketches look similar:
However, if we look at their behavior on a larger interval, then their similarity decreases, but
some of them would still be hard to tell apart just by looking at the shapes of the curves.
Graphs Involving Tangent and Cotangent Functions
EXAMPLE: Graph each function.
(a) y = 2 tan x
(b) y = − tan x
Solution:
(a) To graph y = 2 tan x, we multiply the y-coordinate of each point on the graph of y = tan x
by 2.
(b) The graph of y = − tan x is obtained from that of y = tan x by reflecting in the x-axis.
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EXAMPLE: Graphs of variants of trigonometric curves y = tan 2x + 1 and y = cot 2x + 1 have
π
period and vertical shift 1.
2
EXAMPLE: Graph each function.
(a) y = tan 2x
π
(b) y = tan 2 x −
4
Solution:
(a) The period is π ÷ 2 = π/2 and an appropriate interval is (−π/4, π/4). The endpoints
x = −π/4 and x = π/4 are vertical asymptotes. Thus, we graph one complete period of the
function on (−π/4, π/4). The graph has the same shape as that of the tangent function, but
is shrunk horizontally by a factor of 1/2. We then repeat that portion of the graph to the left
and to the right.
(b) The graph is the same as that in part (a), but it is shifted to the right π/4.
EXAMPLE: Graph each function.
x
(a) y = tan
(b) y = tan πx
2
(c) y = −3 tan 2x
3
EXAMPLE: Graph each function.
x
(b) y = tan πx
(c) y = −3 tan 2x
(a) y = tan
2
Solution:
(a) The period is π ÷ 21 = 2π and an appropriate interval is (−π, π). The endpoints x = −π and
x = π are vertical asymptotes. Thus, we graph one complete period of the function on (−π, π).
The graph has the same shape as that of the tangent function, but is stretched horizontally by
a factor of 2. We then repeat that portion of the graph to the left and to the right.
(b) The period is π ÷π = 1 and an appropriate interval is (−1/2, 1/2). The endpoints x = −1/2
and x = 1/2 are vertical asymptotes. Thus, we graph one complete period of the function on
(−1/2, 1/2). The graph has the same shape as that of the tangent function, but is shrunk
horizontally by a factor of 1/π. We then repeat that portion of the graph to the left and to the
right.
(c) The period is π ÷ 2 = π/2 and an appropriate interval is (−π/4, π/4). The endpoints
x = −π/4 and x = π/4 are vertical asymptotes. Thus, we graph one complete period of the
function on (−π/4, π/4). The graph has the same shape as that of the tangent function, but is
shrunk horizontally by a factor of 1/2, stretched vertically by a factor of 3 and reflected about
the x-axis. We then repeat that portion of the graph to the left and to the right.
EXAMPLE: Graph each function.
π
x
(b) y = 2 cot 3x −
(a) y = 2 cot
3
2
4
(c) y = 3 cot (πx − π)
EXAMPLE: Graph each function.
π
x
(b) y = 2 cot 3x −
(a) y = 2 cot
(c) y = 3 cot (πx − π)
3
2
Solution:
(a) The period is π ÷ 31 = 3π and an appropriate interval is (0, 3π). The endpoints x = 0 and
x = 3π are vertical asymptotes. Thus, we graph one complete period of the function on (0, 3π).
The graph has the same shape as that of the cotangent function, but is stretched horizontally
by a factor of 3 and stretched vertically by a factor of 2. We then repeat that portion of the
graph to the left and to the right.
π
(b) We first put this in the form y = a cot k(x − b) by factoring 3 from the expression 3x − :
2
π
π
y = 2 cot 3x −
= 2 cot 3 x −
2
6
Thus the graph is the same as that of y = 2 cot 3x but is shifted to the right π/6. The period
of y = 2 cot 3x is π ÷ 3 = π/3, and an appropriate interval is (0, π/3). To get the corresponding
interval for the desired graph, we shift this interval to the right π/6. This gives
π π π π π 0+ , +
,
=
6 3
6
6 2
One can get
result in an other way: Since the period of y = cot x is π, the function
theπsame
π
will go through one complete period as 3x − varies from 0 to π.
y = 2 cot 3x −
2
2
Start of period:
End of period:
π
π
3x − = 0
3x − = π
2
2
3π
π
3x =
3x =
2
2
π
π
x=
x=
6
2
Finally, we graph one period in the shape of cotangent on the interval (π/6, π/2) and repeat
that portion of the graph to the left and to the right.
(c) We first put this in the form y = a cot k(x − b) by factoring π from the expression πx − π:
y = 3 cot (πx − π) = 3 cot π (x − 1)
Thus the graph is the same as that of y = 3 cot πx but is shifted to the right 1. The period of
y = 3 cot πx is π ÷ π = 1, and an appropriate interval is (0, 1). Since the length of this interval
and the shift are both 1, we graph one period in the shape of cotangent on the interval (0, 1)
and repeat that portion of the graph to the left and to the right.
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Graphs Involving the Cosecant and Secant Functions
EXAMPLE: Graph each function.
1
π
1
(b) y = csc 2x +
(a) y = csc 2x
2
2
2
Solution:
(a) The period is 2π ÷ 2 = π. An appropriate interval is [0, π], and the asymptotes occur in
this interval whenever sin 2x = 0. So the asymptotes in this interval are x = 0, x = π/2, and
x = π. With this information we sketch on the interval [0, π] a graph with the same general
shape as that of one period of the cosecant function.
(b) We first write
1
π 1
π
y = csc 2x +
= csc 2 x +
2
2
2
4
From this we see that the graph is the same as that in part (a), but shifted to the left π/4.
EXAMPLE: Graph each function.
(a) y = sec 2x
π
(c) y = 2 csc x +
4
1
(b) y = 3 sec x
2
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EXAMPLE: Graph each function.
(a) y = sec 2x
π
(c) y = 2 csc x +
4
1
(b) y = 3 sec x
2
Solution:
(a) The period is 2π÷2 = π. An appropriate interval is [−π/2, π/2], and the asymptotes occur in
this interval wherever cos 2x = 0. Thus, the asymptotes in this interval are x = −π/4, x = π/4.
With this information we sketch on the interval [−π/2, π/2] a graph with the same general shape
as that of one period of the secant function. The complete graph in the Figure below is obtained
by repeating this portion of the graph to the left and to the right.
(b) The period is 2π ÷ 21 = 4π. An appropriate interval is [0, 4π], and the asymptotes occur
in this interval wherever cos 21 x = 0. Thus, the asymptotes in this interval are x = π, x = 3π.
With this information we sketch on the interval [0, 4π] a graph with the same general shape as
that of one period of the secant function. The complete graph in the Figure below is obtained
by repeating this portion of the graph to the left and to the right.
(c) The period is 2π ÷ 1 = 2π. An appropriate interval is
h
πi
π 7π
π
= − ,
0 − , 2π −
4
4
4 4
π
and the asymptotes occur in this interval wherever sin x +
= 0. Thus, the asymptotes in
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this interval are x = π/4, x = 3π/4, and x = 7π/4. With this information we sketch on the
interval [−π/4, 7π/4] a graph with the same general shape as that of one period of the secant
function. The complete graph in the Figure below is obtained by repeating this portion of the
graph to the left and to the right.
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